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Formatted question description: https://leetcode.ca/all/2243.html

# 2243. Calculate Digit Sum of a String (Easy)

You are given a string s consisting of digits and an integer k.

A round can be completed if the length of s is greater than k. In one round, do the following:

1. Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.
2. Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.
3. Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.

Return s after all rounds have been completed.

Example 1:

Input: s = "11111222223", k = 3
Output: "135"
Explanation:
- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".
Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5.
So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round.
- For the second round, we divide s into "346" and "5".
Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5.
So, s becomes "13" + "5" = "135" after second round.
Now, s.length <= k, so we return "135" as the answer.


Example 2:

Input: s = "00000000", k = 3
Output: "000"
Explanation:
We divide s into "000", "000", and "00".
Then we calculate the digit sum of each group: 0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0.
s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".


Constraints:

• 1 <= s.length <= 100
• 2 <= k <= 100
• s consists of digits only.

Similar Questions:

## Solution 1.

• class Solution {
public String digitSum(String s, int k) {
while (s.length() > k) {
int n = s.length();
StringBuilder t = new StringBuilder();
for (int i = 0; i < n; i += k) {
int x = 0;
for (int j = i; j < Math.min(i + k, n); ++j) {
x += s.charAt(j) - '0';
}
t.append(x);
}
s = t.toString();
}
return s;
}
}

• class Solution {
public:
string digitSum(string s, int k) {
while (s.size() > k) {
string t;
int n = s.size();
for (int i = 0; i < n; i += k) {
int x = 0;
for (int j = i; j < min(i + k, n); ++j) {
x += s[j] - '0';
}
t += to_string(x);
}
s = t;
}
return s;
}
};

• class Solution:
def digitSum(self, s: str, k: int) -> str:
while len(s) > k:
t = []
n = len(s)
for i in range(0, n, k):
x = 0
for j in range(i, min(i + k, n)):
x += int(s[j])
t.append(str(x))
s = "".join(t)
return s


• func digitSum(s string, k int) string {
for len(s) > k {
t := &strings.Builder{}
n := len(s)
for i := 0; i < n; i += k {
x := 0
for j := i; j < i+k && j < n; j++ {
x += int(s[j] - '0')
}
t.WriteString(strconv.Itoa(x))
}
s = t.String()
}
return s
}

• function digitSum(s: string, k: number): string {
let ans = [];
while (s.length > k) {
for (let i = 0; i < s.length; i += k) {
let cur = s.slice(i, i + k);
ans.push(cur.split('').reduce((a, c) => a + parseInt(c), 0));
}
s = ans.join('');
ans = [];
}
return s;
}